• Предмет: Алгебра
  • Автор: LadaPolina
  • Вопрос задан 9 лет назад

Вычислите

а) arcs in (-1:2)

б) cos ( 2 arcs in 1:2)

 

            17pi

в)cos   -----   

              3

 

г) tg 600

Ответы

Ответ дал: BuTaC3
0

a)arcsin((-1) / 2) = -0.523598776
arccos(-1) = 3.14159265)

б)

cos(arcsin(-1/2)-arcsin 1)

arcsin(-1/2) = -π/6

arcsin 1 = π/2

cos(arcsin(-1/2)-arcsin 1)=

= cos(-π/6-π/2)=

= cos(-4π/6) =

= cos(2π/3) =

= -1/2

в)sin (17 pi/3) cos (5pi/4) = 1/2{2sin(17pi/3) cos(5pi/4)}
= 1/2{2 sin(6pi - pi/3) cos (pi +pi/4)
= 1/2{ -2sin(pi/3) (- cos(pi/4)}
= 1/2{2sin(pi/3) cos(pi/4)}
= 1/2{ sin(pi/3+pi4) + sin(pi/3- pi/4)}
= 1/2{ sin(7pi/12) +sin(pi/12}}
= 1/2{ sin (105) + sin (15)}
= 1/2{sin(90+15) + sin(15)}
= 1/2{ cos(15) +sin(15)}
= 1/2{cos(45-30) + sin(45-30)}
=1/2[{cos(45)cos (30)+sin(45)sin(30)} + {sin(45)cos(30) - cos(45) sin(30)}]
=1/2{2 (1/sqrt(2))(sqrt(3)/2 )
= 1/2{sqrt(3/2)} 

г)

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